SirMinus AMA

[Deleted User]

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[Balls]

There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original’s minus one, and whose last two digits are the same and equal to the half of the original’s.
Find that number.

[Foxirius2205]

WOW ?

[Deleted User]

WOW ?

ye

[Deleted User]

There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original’s minus one, and whose last two digits are the same and equal to the half of the original’s.
Find that number.

1.5 kilos of meth

[User 12049]

There is a two-digit number whose digits are the same, and has got the following property: When squared, it produces a four-digit number, whose first two digits are the same and equal to the original’s minus one, and whose last two digits are the same and equal to the half of the original’s.
Find that number.

Firstly, let us assume that the digit we are looking for is
x
Then, the whole number becomes
(xx)−(xx)−
Whose value is equal to
(xx)−=10x+x=11x(xx)−=10x+x=11x
Subsequently, we raise the number to the second power, and becomes
(xx)−=(11x)2=121x2(xx)−=(11x)2=121x2
Now that our number is raised to the second power, we proceed to the second part. If we assume that our original number is
(xx)−(xx)−
It is clear that the number produced after the squaring is equal to
((x−1)(x−1)x2x2)−((x−1)(x−1)x2x2)−
And its value is
1000(x−1)+100(x−1)+10x2+1x21000(x−1)+100(x−1)+10x2+1x2
By applying the distributive property, the number becomes
1000x−1000+100x−100+10x2+x21000x−1000+100x−100+10x2+x2
To add the terms together, it is necessary to convert all amounts to like quantities. This can be achieved by multiplying each term (apart from the two last) by 2/2
1000x⋅22−1000⋅22+100x⋅22−100⋅22+10x2+x21000x⋅22−1000⋅22+100x⋅22−100⋅22+10x2+x2
After, the number becomes
2000x2−20002+200x2−2002+10x2+x22000x2−20002+200x2−2002+10x2+x2
We add the terms together, making the number
2000x−2000+200x−200+10x+x22000x−2000+200x−200+10x+x2

Adding the numerator’s terms we get
2211x−220022211x−22002
Now, after finishing with both our numbers, we equate them
121x2=2211x−22002121x2=2211x−22002
We multiply each side by 2, and the equation becomes
242x2=2211x−2200242x2=2211x−2200
We move all terms to the left, making the second degree trinomial equal to zero
242x2−2211x+2200=0242x2−2211x+2200=0
After, we can factor our expression, just to make things easier.
We notice that 242, 2211 and 2200 have a common factor of 11.
So, we can factor the whole expression into
11(22x2−201x+200)=011(22x2−201x+200)=0
We divide both sides by eleven, and we end up with
22x2−201x+200=022x2−201x+200=0
Now, it is clear that we have got a second degree equation, which is equal to zero. Our first task is to calculate the discriminant, which is given by the formula
D=b2−4acD=b2−4ac, where a=22, b=201 and c=200
We replace the variables with the values, and the discriminant becomes equal to
D=22801
Now, we need to figure out whether the equation has got real roots or not.
We have the following possibilities:
b2−4ac>0⇒b2−4ac>0⇒ There are two real roots.
The solution x of the second degree equation is given by the formula
x1,2=−b±b2−4ac√2ax1,2=−b±b2−4ac2a
x1,2=201±15144x1,2=201±15144
x1=201+15144=35244=8x1=201+15144=35244=8
x2=201−15144=5044=1.13636363636...x2=201−15144=5044=1.13636363636...
Finally, we come to the point where we have to choose between the two possible solutions for x. Since the value we need concerns a digit, it has to be an integer.
8∈Z8∈Z True
1.13636363636...∈Z1.13636363636...∈Z False

Hence, we come to the conclusion that x=8.
(xx)−=88

[BlueHeartViper]

You like memes?

[Deleted User]

You like memes?

Oui oui.

[trxsh]

Can we ERP in the Nexus Elevator?